【线段树专题】poj2104

来源：互联网发布：淘宝实木家具编辑：程序博客网时间：2024/05/18 00:58

这个题已经被花式A了

因为要练习线段树，所以就用函数式线段树写了这个题，

函数式运用了一种可持久化的思想，在修改的同时记录修改以前的样子，

函数式线段树的话就是在每次进行插入操作的时候在上一次线段树的基础上新建一条路径，其余节点都用原来线段树的节点

感觉现在写起来比以前要轻松很多了。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<algorithm>using namespace std;int root[200100],a[200100],b[200100],tot,n,q,tail;struct rec{    int l,r,lch,rch,sum;}tree[4001000];void maketree(int l,int r){    int now=++tot;    tree[now].l=l;    tree[now].r=r;    if (l==r) return;    int mid=(l+r)>>1;    tree[now].lch=tot+1;maketree(l,mid);    tree[now].rch=tot+1;maketree(mid+1,r);}void insert(int x1,int x2,int w){    int now=++tot;    tree[tot].sum=tree[x1].sum;    tree[tot].l=tree[x1].l;    tree[tot].r=tree[x1].r;    tree[tot].lch=tree[x1].lch;    tree[tot].rch=tree[x1].rch;    if (tree[tot].l==tree[tot].r)    {        tree[tot].sum++;        return;    }    int mid=(tree[x1].l+tree[x1].r)>>1;    if (w<=mid)     {        tree[tot].lch=tot+1;        insert(tree[x1].lch,tot+1,w);    }    else    {        tree[tot].rch=tot+1;        insert(tree[x1].rch,tot+1,w);    }    tree[now].sum++;}int find_ans(int x1,int x2,int k){    if (tree[x1].l==tree[x1].r)        return a[tree[x1].l];    if (tree[tree[x2].lch].sum-tree[tree[x1].lch].sum>=k)        return find_ans(tree[x1].lch,tree[x2].lch,k);    else        return find_ans(tree[x1].rch,tree[x2].rch,k-(tree[tree[x2].lch].sum-tree[tree[x1].lch].sum));}int main(){    freopen("poj2104.in","r",stdin);    tot=0;    scanf("%d%d",&n,&q);    for (int i=1;i<=n;i++)    {        scanf("%d",&a[i]);        b[i]=a[i];    }    sort(a+1,a+n+1);    tail=unique(a+1,a+n+1)-a-1;    maketree(1,tail);    root[0]=1;    int w;    for (int i=1;i<=n;i++)    {        root[i]=tot+1;        w=lower_bound(a+1,a+tail+1,b[i])-a;        insert(root[i-1],root[i],w);    }    int l,r,k,ans;    for (int i=1;i<=q;i++)    {        scanf("%d%d%d",&l,&r,&k);        ans=find_ans(root[l-1],root[r],k);        printf("%d\n",ans);    }}

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